I'm not sure if these details matter, but anyway for this particular case, consider a compact abelian group $G$ with operation $\cdot$, a Haar measure $\mu$ on it and $f$ a non-trivial character on $G$. (I'm looking at this question: The integral of a character is $0$)

I've seen stated that for $x,y\in G$,

$$\int_G f(x)\>d\mu(x)=\int_G f(x\cdot y)\>d\mu(x\cdot y).$$

Is this true? If so why? What's the general theory behind it?

  • $\begingroup$ Hint: recall the definition of (right/left) Haar measure (in particular, recall that $\mu(xA)=\mu(A)$). First show that the equation holds for simple functions, and then use a density argument to prove the general case. $\endgroup$ – Manuel Norman May 24 at 19:59
  • $\begingroup$ @ManuelNorman I think the part that's tripping me up is if we know $\mu(xA)=\mu(A)$, why does that hold for $d\mu$? $\endgroup$ – J.Smith May 24 at 20:06
  • $\begingroup$ This is usually a choice of notation: for instance, you could also simply write $d \mu$, or $\mu (dx)$; sometimes, it is not even explicitely written. $\endgroup$ – Manuel Norman May 24 at 20:12

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